No, the set of all subsets of the integer is not countable. Since $\mathbb{Z}$ has the same cardinality as $\mathbb{N}$, it suffice to consider all subsets of $\mathbb{N}$.
For each subset $X$ of $\mathbb{N}$ consider the characteristic function $\chi_X$ defined by
$\chi_X(z) = \begin{cases}
1 & \quad z \in X \\
0 & \quad z \notin X
\end{cases}$
In this way you associate injectively and subjectively each subset $X$ of $\mathbb{N}$ with a function in $2^{\mathbb{N}}$. $2^\mathbb{N}$ has cardinality strictly larger than $\mathbb{N}$. This is proved by the typical Cantor diagonalization argument.
Also, Cantor Diagonalization and the function I wrote above can be used to show more generally that the set of all subsets of a given set has cardinality strictly greater than the given set.
In response to comment :
You can think of a function from $\mathbb{N} \rightarrow 2$ a infinite binary strings of $0$'s and $1$'s. Assume that $2^{\mathbb{N}}$ is countable. That is there is a bijection $\sigma$ from $\mathbb{N}$ to $2^\mathbb{N}$. Then define the function $h : \mathbb{N} \rightarrow 2$ as follows
$h(n) = \begin{cases}
1 & \quad (\sigma(n))(n) = 0 \\
0 & \quad (\sigma(n))(n) = 1
\end{cases}$
Informally, this is the familiar argument, form a new binary string by going down the diagonal and switching $0$ for $1$ and $1$ for $0$. Now this is a a perfectly good binary string hence it down appear as $\sigma(k)$ for some $k$ if $\sigma$ is indeed a bijection. However, it can not be $\sigma(k)$ for any $k$ since it differs from $\sigma(k)$ in at least the $k^\text{th}$ entry.
I hope this helps.
This fact, in the generality presented here, is equivalent to the axiom of choice. So a simple injection from $\cal I$ into $I$ cannot be written out explicitly when you're taking the whole generality.
Let's limit ourselves to sets that can be well-ordered, or rather to ordinals, or rather to initial ordinals (since we care about cardinality). Assuming the axiom of choice every set is equipotent with an initial ordinal, so that's fine.
If $\omega_\alpha$ is an initial ordinal, then every finite set can be thought of as a finite sequence. The knee-jerk reaction is to look at the ordinal $(\omega_\alpha)^\omega$, here the exponent is an ordinal exponentiation. That's absolutely fine, and we can prove that if $\alpha$ is an infinite ordinal then $\alpha$ and $\alpha^\omega$ have the same cardinality (again, ordinal exponentiation!)
But I want to go to a different route. Let's look at $\omega^{\omega_\alpha}$. We can prove, quite easily depending on the definitions and theorems at hand, that this ordinal is $\omega_\alpha$ itself. So it is enough to find an injection from the finite subsets of $\omega_\alpha$ into $\omega^{\omega_\alpha}$.
A direct definition of $\alpha^\beta$ is the order type of finite partial functions from $\beta$ to $\alpha$ which are decreasing. But this gives us a dead giveaway for an injection: given a finite set $J\subseteq\omega_\alpha$, write it in increasing order as $\{\xi_1,\dots,\xi_n\}$, and then simply map $\xi_i$ to $i$.
Therefore we found an injection from $\cal I$ to $\omega^{\omega_\alpha}$ as wanted.
Best Answer
$A$ has $a^\omega$ countably infinite subsets, and there’s not much more that you can say unless you know something about the cardinal $a$. For example, if $2\le a\le 2^\omega=\mathfrak c$, then $a^\omega=2^\omega$. If $\operatorname{cf}a=\omega$, i.e., if $a$ has cofinality $\omega$, then $a^\omega>a$.